Đáp án:
Giải thích các bước giải:
Bài 2:
Ta có: \(\dfrac{a^2+b^2+c^2}{3}-\left(\dfrac{a+b+c}{3}\right)^2\)
\(=\dfrac{a^2+b^2+c^2}{3}-\dfrac{\left(a+b+c\right)^2}{3^2}\)
\(=\dfrac{3a^2+3b^2+3c^2}{9}-\dfrac{a^2+b^2+c^2+2ab+2ac+2bc}{9}\)
\(=\dfrac{2a^2+2b^2+2c^2+2ab+2ac+2bc}{9}\)
\(=\dfrac{\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)}{9}\)
\(=\dfrac{\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2}{9}\)
Vì \(\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\ge0\)
\(\Rightarrow\dfrac{\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2}{9}\ge0\)
\(⇔ \dfrac{a^2+b^2+c^2}{3}-\left(\dfrac{a+b+c}{3}\right)^2\ge0\)
hay \(\dfrac{a^2+b^2+c^2}{3} \ge \left(\dfrac{a+b+c}{3}\right)^2\)
`⇒` đpcm
